Download presentation

Presentation is loading. Please wait.

Published byJessica Boyle Modified over 5 years ago

2
To help students visualize abstract concepts Introduce a new topic; students then can discover the algebra rules instead of being told by the teacher Reinforce a topic for a struggling student

3
Positive TilesNegative Tiles 11 1 1 x x X x x X X² 11 X

4
Addition and Subtractions of Integers Distributive Property Combining Like Terms Solving Equations Multiplying Binominals Factoring Polynomials

5
3 + ( -5 ) = ? Additions means Combine Use the zero property to cancel/take away a pair of blue and red tiles Left with 2 -1 tiles Answer = -2 1 1 1 1 1 1 1 1

6
5 – (-2) = ? Start with 5 take away Answer = 7 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Now you can take away 2 -1 tiles Add a zero pair in order to be able to take away 2 - 1 tiles

7
3 (2x + 1) is equivalent to 6x + 3 3 (2x + 1) {using repeated addition) Rearrange the tiles XX 1 XX XX 1 1 X X X X X X 1 1 1 + +

8
(x² - 2x -3) – (2x² + x – 2) = ? X² X X 1 1 1 Subtraction would be represented by adding the opposite of each term in parenthesis X² X 1 1 Cross out all zero pairs, what you have left over is your answer Answer = -x² – 3x – 1 +

9
2x -2 = 4 X X 11 = 1111

10
Add two tiles to the left to make a zero pair * To keep scale balanced - do the same to both sides * X X 11 = 1111 1111

11
2x -2 = 4 Arrange the tiles into groups Answer: X = 3 1 1 X = X 1 11 1

12
(x + 3) (x + 2) X111 1 X 1 L W

13
X111 1 X 1 X² XXX X X 111 111 Answer = x² + 5x + 6 Fill in the space so that lines between tiles are continuous

14
2x² + 5x + 3 X² X X XX X111 First fill in the x² tiles and 1 tiles Then arrange the x tiles to match

15
2x² + 5x + 3 X² X X XX X111 111 1 XX X x + 1 2x + 3 Answer = (2x +3) ( x + 1)

16
Teaching the rules of Algebra Tiles Multiplying a Polynomial by a Binomial More difficult using negative number for: Distributing Factoring Multiplying two binomials (It is possible but students might struggle)

17
Visualizing an abstract concept Students generating rules Practice with basic problems then have students find patterns to then apply to more challenging problems Another tool in the tool box

Similar presentations

OK

EXAMPLE 4 Solve proportions SOLUTION a. 5 10 x 16 = Multiply. Divide each side by 10. a. 5 10 x 16 = = 10 x5 16 = 10 x80 = x8 Write original proportion.

EXAMPLE 4 Solve proportions SOLUTION a. 5 10 x 16 = Multiply. Divide each side by 10. a. 5 10 x 16 = = 10 x5 16 = 10 x80 = x8 Write original proportion.

© 2019 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google